A320926 Concatenation of successive segments generated by the morphism {0 -> {0, 0, 1}, 1 -> {0}}, starting with 0.

0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1

Offset

1

Comments

Length of n-th segment: A001333(n), for n >= 1.

Positions of 1: A184118.

From Michel Dekking, Aug 10 2022: (Start)

This sequence is a morphic sequence, i.e., (a(n)) is a letter to letter projection of a fixed point of a morphism tau.

The alphabet is {0,1,2}, and tau is given by

tau(0) = 001, tau(1) = 0, tau(2) = 2001.

The letter-to-letter map is given by 0->0, 1->1, 2->0.

One proves easily with induction that the fixed point of tau starting with 2 is mapped to (a(n)) by the letter to letter map. (End)

From Michel Dekking, Aug 13 2022: (Start)

This sequence is constructed from the Pell morphism pi: 0->001, 1->0, which has fixed point the Pell word A171588.

The Pell word is a Sturmian sequence s(alpha, rho) with slope alpha = 1 - 1/sqrt(2) and intercept rho 1 - 1/sqrt(2) (See comments of A171588).

By definition,

(a(n)) = lim_{k->oo} 0 pi(0) pi^2(0) ... pi^k(0).

So any word that occurs in the Pell word A171588, occurs in (a(n)).

Conversely, any word occurring in (a(n)) occurs in the Pell word. This one proves by showing by induction on k that the word 0 1 0 pi(0) pi^2(0) ... pi^k(0) occurs in the Pell word.

Conclusion: (a(n)) and the Pell word generate the same language. But then they also have the same subword complexity function, which is p(n) = n+1 for the Pell word.

But a word has subword complexity function p(n) = n+1 if and only if it is Sturmian, so (a(n)) is a Sturmian sequence (a(n)) = s(alpha*, rho*) for some alpha* and rho*.

It follows directly from Lothaire Proposition 2.2.18 that alpha* = alpha.

Conjecture: rho* = 0.085.... (End)

It can be proved that rho* = 3/2 - sqrt(2) = 0.085786437626904951... - Michel Dekking, Sep 03 2022

Links

Clark Kimberling, Table of n, a(n) for n = 1..20000

M. Lothaire, Algebraic combinatorics on words, Cambridge University Press. Online publication date: April 2013; Print publication year: 2002.

Example

First 4 segments:
  {0},
  {0, 0, 1},
  {0, 0, 1, 0, 0, 1, 0},
  {0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1}

Mathematica

s[n_] := Nest[Flatten[# /. {0 -> {0, 0, 1}, 1 -> {0}}] &, {0}, n];  (* Pell morphism, A171588 *)
t = Table[s[n], {n, 0, 8}];  Take[t, 5]  (* successive segments of morphism s *)
u = Flatten[t]; Take[u, 200]  (* A320926 *)

Crossrefs

Cf. A171588, A184118, A320927.

Sequence in context: A288372 A354321 A072786 * A144597 A125117 A144603

Adjacent sequences: A320923 A320924 A320925 * A320927 A320928 A320929

Keyword

nonn,easy

Author

Clark Kimberling, Oct 24 2018

Status

approved